Question: The following equation is true for all real values of $y$ for which the expression on the left is defined, and $C$ is a polynomial expression. $\dfrac{y^2+7y-18}{C} \cdot \dfrac{14y^2}{7y^2-14y}=1$ What is $C$ ? $C=$
Solution: The left side of the equation is a product of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting product on the left side should cancel out completely. In order to solve for $C$, let's multiply the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The numerator, $y^2+7y-18$, of the first expression can be factored as $(y+9)(y-2)$ using the sum-product pattern. The numerator, $14y^2$, of the second expression cannot be factored any further. The denominator, $7y^2-14y$, of the second expression can be factored as $7y(y-2)$ by factoring out $7y$. Now the product looks as follows: $\dfrac{(y+9)(y-2)}{C} \cdot \dfrac{14y^2}{7y(y-2)}$ To find the product of two rational expressions, we multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{(y+9)(y-2)}{C} \cdot \dfrac{14y^2}{7y(y-2)} \\\\\\ &= \dfrac{(y+9)(y-2) \cdot 14y^2}{C \cdot 7y(y-2)} &\text{Multiply across.}\\\\\\ &= \dfrac{(y+9){\cancel{(y-2)}} \cdot {\cancel{7}} \cdot 2 \cdot {\cancel{y}} \cdot y}{C \cdot {\cancel{7}} {\cancel{y}} {\cancel{(y-2)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{2y(y+9)}{C} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{2y(y+9)}{C}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $C=2y(y+9)$, which is equivalent to $2y^2+18y$.